690 reputation
616
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location United Kingdom
age
visits member for 3 years, 2 months
seen 14 hours ago

Manual worker. Live in England. Formal scientific education "peaked" many years ago when I failed maths and chemistry A-level and scraped a bare pass in physics. Still enjoy pottering about in maths and physics.


Jan
22
comment Is this interpretation of $\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$ correct?
I cut a few corners in my question and have now been found out. My original calculation of the probability $P$ of finding the electron in a sphere radius $x$ was along the lines of $$P=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{x}\left|\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}\right|^{2}\left(r^{2}\sin\theta drd\theta d\phi\right)=\frac{4\pi}{\pi a^{3}}\int_{0}^{x}r^{2}e^{-2r/a}dr.$$ continued ...
Jan
21
asked Is this interpretation of $\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$ correct?
Jan
15
accepted Calculation of spherical Bessel functions - meaning of $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$
Jan
15
comment Calculation of spherical Bessel functions - meaning of $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$
Thanks. So, in this case, you'd simply take the derivative, then multiply by 1/x, then take the derivative again and then multiply by 1/x? Afraid I'm not too familiar with operators.
Jan
15
asked Calculation of spherical Bessel functions - meaning of $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$
Dec
31
accepted Don't understand the integral over the square of the Dirac delta function
Dec
31
comment Don't understand the integral over the square of the Dirac delta function
To me, that seems even clearer (in so far as anything involving this bizarre function can be clear). Apologies twistor59, I've now accepted this answer.
Dec
31
comment Don't understand the integral over the square of the Dirac delta function
That's plausible enough for me. Out of interest, why isn't this a proof?
Dec
30
comment Don't understand the integral over the square of the Dirac delta function
This is a little advanced for me as I'm not familiar with Heaviside step functions. Twistor59's answer is more my level though I'm still trying to think it through.
Dec
30
asked Don't understand the integral over the square of the Dirac delta function
Dec
24
awarded  Critic
Jun
17
awarded  Yearling
May
7
comment Zero divergence of energy-momentum tensor and gravitational energy
apologies for being a bit slow on the uptake. This stuff would be a lot easier if (a) I was smarter and (b) I had a degree in physics. I confess I've only just today found out what a continuity equation is! I've also just read a reasonably understandable on-line piece by Weiss and Baez titled, "Is energy conserved in general relativity". Their answer is an unambiguous, "In special cases, yes. In general — it depends on what you mean by 'energy', and what you mean by 'conserved'." Glad that's sorted out.
May
7
accepted Zero divergence of energy-momentum tensor and gravitational energy
May
6
comment Zero divergence of energy-momentum tensor and gravitational energy
Is it possible to describe, in absolute baby steps why that implication is true?
May
6
comment Zero divergence of energy-momentum tensor and gravitational energy
thanks. Any chance of more slowly and even simplier, with no reference to Lagrangians? I have a rudimentary understanding of the divergence theorem, connection coefficients and covariant derivatives. My understanding of divergence is at the level of sources and sinks of a liquid sloshing about in a container - if no liquid is entering or leaving the container, the divergence equals zero. I'm trying to relate that simple picture to $\nabla_{\mu}T^{\mu\nu}=0$ and why that implies the conservation of energy in flat but not in curved spacetime.
May
5
comment Zero divergence of energy-momentum tensor and gravitational energy
of the rhs (the EMT) doesn't imply the conservation of energy and momentum. If you put the Einstein tensor on the rhs (ie, the lhs equals zero), would the zero divergence of the rhs then imply total conservation of energy and momentum?
May
5
comment Zero divergence of energy-momentum tensor and gravitational energy
in some way from the rhs, hence my confusion. I didn't know that (as you say) “the curvature of spacetime is ONLY dictated by the density of the non-gravitational energy density T or EMT”. So, thankfully, all the calculations I've struggled through re Schwarzschild and cosmology are still valid because only the EMT (and not gravitational energy) curves spacetime? I'm still puzzled about why zero divergence of the rhs (the EMT) doesn't imply the conservation of energy and momentum. If you put the Einstein tensor on the rhs (ie, the lhs equals zero), would the zero divergence
May
5
comment Zero divergence of energy-momentum tensor and gravitational energy
Oh dear. I think I've got this seriously wrong. Please bear with me. I thought the rhs of the field equations (ie the EMT) described the total energy-momentum of a system (which I think is correct). I thought the lhs was a description of the curvature of spacetime caused by the rhs (which I think also is correct). But I never thought of the lhs as describing gravitational energy. (Thinking about it, both sides must of course have the same units, so if the rhs describes energy so must the lhs). I thought that the “additional source of energy”, ie gravitational energy, was missing
May
5
comment Zero divergence of energy-momentum tensor and gravitational energy
thanks. I did try to read the link but it was a little over my head (my level - I've no idea what Hamiltonians or Lagrangians are!). I've been trying to understand the Schwarzshild metric and relativistic cosmology both of which are based on a particular energy-momentum tensor (zero and perfect fluid). I still don't see how that can be valid if the EMT doesn't include gravitational energy. I sense you answering my question in your final paragraph ("very-long-distance effective description") but, sorry, don't understand. Please don't worry about making your answer too simple!